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2 years ago

Which answer describes the measure of the amount of matter ?

2 answers:

7 0

A) molar mass.

Molar mass is the mass of one mole of a pure substance. So basically the measure of the amount of matter

Cerrena [4.2K]2 years ago

7 0

Answer: B. mole/s

Explanation:

Molar mass is defined as the mass of 1 mole of the compound. It is calculated by adding the masses of all the elements present in the compound.

A mole is defined as the amount of substance that contains Avogardro number of the substance. Avogadro's number is given by $6.022\times 10^{23}$ .

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles. Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atmosphere respectively.

Pressure of the gas is defined as the force exerted by the particles on the walls of the container. It is expressed in various terms like 'mmHg', 'atm', 'kiloPascals' etc..

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Consider the reaction pathway graph below.

Alex17521 [72]

Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

$Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ$

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!

6 0

3 years ago

How many milliliters of a 8.20 m licl solution contain 18.4 g of licl?

Dovator [93]

Your final answer is that you will need 52.9mL of the 8.20 M of LiCl

3 0

2 years ago

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

lukranit [14]

I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.

5 0

3 years ago

Investigate Weathering and Erosion 1. Describe at least one part of the experimental procedure that you thought was essential to

zvonat [6]

Answer:

Controlling the environment is the most key procedures for getting good results.

Explanation:

The control environment for an experiment is the essential part for getting good results. In control environment, there is no or less chances of disruption

from the external environment which can cause the results of the data more acceptable. So the scientists prefers laboratory for performing experiment as compared to outer environment. So in my opinion for getting better results, the control environment is the most necessary experimental procedure.

4 0

2 years ago

Read 2 more answers